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### Network # Autocorrecting multimeter?

L

#### LabMonkey

Jan 1, 1970
0
Dear Newgroup,

I came across an interesting phenomenon:

In the lab we placed a "100M" Ohm resistor (whose resistance we
couldn't measure directly with the multimeter) in series with a 10MOhm
resistor (which we could measure). Building a basic voltage divider (3V
battery) one should get 0.3V on the 10M resistor.

Now the internal resistance of both multimeters we have is 10M (Fluke
175, Amprobe 18-A), so we are placing two 10M in parallel. Doing the
basics: the equivalent resistor combination is then 5M (sorry if that
was insulting which would cause a 0.15V drop to be measured.

Instead, on the Amprobe 18-A I read the 0.3V, while on the Fluke I get
0.15V. So apparently the Amprobe somehow realizes it is changing the
voltage and corrects for it.

Has anyone ever come across this before?

Any ideas on how the meter does this?

LabMonkey

E

#### eehinjor

Jan 1, 1970
0
I have done such experiment.this depends on the precesion you want,and
the multimeter's internal resistance.

D

#### Dan Hollands

Jan 1, 1970
0
I doubt there is any autocorrecting going on - I suspect the input impedance
is just much higher than speced - may be spec is really "greater then 10M"

Dan

--
Dan Hollands
1120 S Creek Dr
Webster NY 14580
585-872-2606
[email protected]
www.QuickScoreRace.com

L

#### LabMonkey

Jan 1, 1970
0
Once we know the resistance of the "100M" resistor we then place it in
series with the multimeter and the battery. In essence we are replacing
the known 10M resistor with the unknown internal resistance of the
multimeter.

For both multimeters we are getting a votlage reading of 0.3V - which
means 10M internal resistance.

As for the precision of the reading, both multimeters give us three
digits after the decimal point. So it isn't a problem of rounding.

Any ideas of how to verify the internal resistance of the meters
another way?

LabMonkey

J

#### John Fields

Jan 1, 1970
0
Once we know the resistance of the "100M" resistor we then place it in
series with the multimeter and the battery. In essence we are replacing
the known 10M resistor with the unknown internal resistance of the
multimeter.

For both multimeters we are getting a votlage reading of 0.3V - which
means 10M internal resistance.

As for the precision of the reading, both multimeters give us three
digits after the decimal point. So it isn't a problem of rounding.

Any ideas of how to verify the internal resistance of the meters
another way?

J

#### John Fields

Jan 1, 1970
0
Dear Newgroup,

I came across an interesting phenomenon:

In the lab we placed a "100M" Ohm resistor (whose resistance we
couldn't measure directly with the multimeter) in series with a 10MOhm
resistor (which we could measure). Building a basic voltage divider (3V
battery) one should get 0.3V on the 10M resistor.

---

E1
|
[R1]
|
+---E2
|
[R2]
|
GND

E1R2 3V * 10M
E2 = ------- = ------------ = 0.2727... V
R1+R2 100M + 10M

---
Now the internal resistance of both multimeters we have is 10M (Fluke
175, Amprobe 18-A), so we are placing two 10M in parallel. Doing the
basics: the equivalent resistor combination is then 5M (sorry if that
was insulting which would cause a 0.15V drop to be measured.

E1
|
[R1]
|
+---E2
|
[R2]
|
GND

E1R2 3V * 5M
E2 = ------- = ------------ ~ 0.1428V
R1+R2 100M + 5M

Instead, on the Amprobe 18-A I read the 0.3V, while on the Fluke I get
0.15V. So apparently the Amprobe somehow realizes it is changing the
voltage and corrects for it.

Has anyone ever come across this before?

---
Dunno, but I haven't.
---
Any ideas on how the meter does this?

---
Nope, but if it was me I'd have it do something like this:

Vbat
|
Vin+<---+ |
| |
[R1] [VR1]
| |
E1--+--[R3]--+--E2
| |
[R2] [R4]
| |
Vin-<---+--------+
<--I2

Where R1 and R2 are 5 megohms each, giving me the 10 megohms input
resistance I need for the meter, but upset by R3 + R4 being in
parallel with R2. After connecting to Vin+ and Vin- and and
detecting current in R3 and R4 I'd turn on VR1 (a FET, probably)
and ramp it through its resistance range until the current in R3
went to zero (or until the currents in VR1 and R4 were equal). At
that point the bridge would be balanced, the input resistance would
be R1 + R2, and I'd measure the current in R4 (or VR1) and
calculate:

E2 = I2R4 = E1

L

#### LabMonkey

Jan 1, 1970
0
Thanks for all your help. I will try and find the parts to build the
circuit and compare it with the capabilities of the meter.

J

#### John Fields

Jan 1, 1970
0
Thanks for all your help. I will try and find the parts to build the
circuit and compare it with the capabilities of the meter.

J

#### John Larkin

Jan 1, 1970
0
Dear Newgroup,

I came across an interesting phenomenon:

In the lab we placed a "100M" Ohm resistor (whose resistance we
couldn't measure directly with the multimeter) in series with a 10MOhm
resistor (which we could measure). Building a basic voltage divider (3V
battery) one should get 0.3V on the 10M resistor.

Now the internal resistance of both multimeters we have is 10M (Fluke
175, Amprobe 18-A), so we are placing two 10M in parallel. Doing the
basics: the equivalent resistor combination is then 5M (sorry if that
was insulting which would cause a 0.15V drop to be measured.

Instead, on the Amprobe 18-A I read the 0.3V, while on the Fluke I get
0.15V. So apparently the Amprobe somehow realizes it is changing the
voltage and corrects for it.

Has anyone ever come across this before?

Any ideas on how the meter does this?

LabMonkey

Sounds like the Amprobe actually has near-infinite resistance on its
low range; that's not unheard of. It does *not* correct for source
resistance!

Put the Fluke and the Amprobe in series, measure the battery, and tell
us what they report.

John

J

#### Jasen Betts

Jan 1, 1970
0
Dear Newgroup,

I came across an interesting phenomenon:

In the lab we placed a "100M" Ohm resistor (whose resistance we
couldn't measure directly with the multimeter) in series with a 10MOhm
resistor (which we could measure). Building a basic voltage divider (3V
battery) one should get 0.3V on the 10M resistor.

Now the internal resistance of both multimeters we have is 10M (Fluke
175, Amprobe 18-A), so we are placing two 10M in parallel. Doing the
basics: the equivalent resistor combination is then 5M (sorry if that
was insulting which would cause a 0.15V drop to be measured.

Instead, on the Amprobe 18-A I read the 0.3V, while on the Fluke I get
0.15V. So apparently the Amprobe somehow realizes it is changing the
voltage and corrects for it.

What figures do you get if connect both voltmeters in series across the 3V
supply?

I get the feeling that atleast one of the meters isnt 10M resistance.

for measuring the "100M" resistor how about connecting it in series with the
meter and a 1M resistor and measuring how much current it passes when a
known voltage is applied?

Bye.
Jasen

L

#### LabMonkey

Jan 1, 1970
0
You were right (although it is a little trickier Placing a 3V battery in series with the Fluke and Amprobe meter I get
1.5V on both IF I have the Amprobe meter on a range other than mV. If
instead I have the Amprobe on the mV range the votlage it displays is
naturally "Ol" while on the Fluke I read 0.095V.

Doing the math I get 300M internal resistance for the Amprobe on the mV
range and 10M on any other range.

Oh the joy of undocumented features.

Thanks for all the input from all of you!

L

#### LabMonkey

Jan 1, 1970
0
Actually I spoke too soon.

To answer your first question: when having the Amprobe meter on the
Volt range both meters read 1.5V which is what you get when the
internal resistance are the same. If instead I place the Amprobe meter
on the mV range it reads Ol and the other meter reads 0.095V instead of
1.5V. This indicates that on the mV range the Amprobe meter uses a
different resistance than on the Volt range. From the initial battery
voltage, the votlage drop on the Fluke meter (which was in series) and
the internal resistance of the Fluke (10M) I got 0.3GOhms.

I will try out your suggestion once I find a 1MOhm resistor.

Now onto the latest problem. When I place the Amprobe meter in series
with the 100M resistor and apply the 3V battery to both I get 0.3V drop
displayed on the meter when it is in the Volt range. If I place it in
the mV range I yet again read Ol. But having an electrometer I used
that to measure the votlage drop on the 100M resistor, which is 0.2V.
Solving again the simple votlage divider equation I now get out an
internal resistance of 1.4GOhms for the Amprobe meter (instead of the
0.3G).

I will do it again tomorrow just to verify the result.

J

#### John Fields

Jan 1, 1970
0
Actually I spoke too soon.

To answer your first question: when having the Amprobe meter on the
Volt range both meters read 1.5V which is what you get when the
internal resistance are the same. If instead I place the Amprobe meter
on the mV range it reads Ol and the other meter reads 0.095V instead of
1.5V. This indicates that on the mV range the Amprobe meter uses a
different resistance than on the Volt range. From the initial battery
voltage, the votlage drop on the Fluke meter (which was in series) and
the internal resistance of the Fluke (10M) I got 0.3GOhms.

I will try out your suggestion once I find a 1MOhm resistor.

Now onto the latest problem. When I place the Amprobe meter in series
with the 100M resistor and apply the 3V battery to both I get 0.3V drop
displayed on the meter when it is in the Volt range. If I place it in
the mV range I yet again read Ol. But having an electrometer I used
that to measure the votlage drop on the 100M resistor, which is 0.2V.
Solving again the simple votlage divider equation I now get out an
internal resistance of 1.4GOhms for the Amprobe meter (instead of the
0.3G).

I will do it again tomorrow just to verify the result.

---
Try this:

3V E1
|
[100M]R1
|
+------+--E2
| |
[Rx] [Rm]
| |
+------+
|
GND

With Rm (your meter) on the millivolt range, adjust Rx until you get
some arbitrary voltage reading (E2) on the meter. Record the value
of Rx and E2.

Now solve for the parallel combination of Rx and Rm:

E2 R1
Rt = -------
E1-E2

Now, knowing Rx, solve for Rm:

Rt R1
Rm = -------
R1-Rt

J

#### John Fields

Jan 1, 1970
0
Try this:

3V E1
|
[100M]R1
|
+------+--E2
| |
[Rx] [Rm]
| |
+------+
|
GND

With Rm (your meter) on the millivolt range, adjust Rx until you get
some arbitrary voltage reading (E2) on the meter. Record the value
of Rx and E2.

Now solve for the parallel combination of Rx and Rm:

E2 R1
Rt = -------
E1-E2

Now, knowing Rx, solve for Rm:

Rt R1
Rm = -------
R1-Rt

---
Oops...

Rx Rt
Rm = -------
Rx-Rt

J

#### John Larkin

Jan 1, 1970
0
Actually I spoke too soon.

To answer your first question: when having the Amprobe meter on the
Volt range both meters read 1.5V which is what you get when the
internal resistance are the same. If instead I place the Amprobe meter
on the mV range it reads Ol and the other meter reads 0.095V instead of
1.5V. This indicates that on the mV range the Amprobe meter uses a
different resistance than on the Volt range. From the initial battery
voltage, the votlage drop on the Fluke meter (which was in series) and
the internal resistance of the Fluke (10M) I got 0.3GOhms.

I will try out your suggestion once I find a 1MOhm resistor.

Now onto the latest problem. When I place the Amprobe meter in series
with the 100M resistor and apply the 3V battery to both I get 0.3V drop
displayed on the meter when it is in the Volt range. If I place it in
the mV range I yet again read Ol. But having an electrometer I used
that to measure the votlage drop on the 100M resistor, which is 0.2V.
Solving again the simple votlage divider equation I now get out an
internal resistance of 1.4GOhms for the Amprobe meter (instead of the
0.3G).

I will do it again tomorrow just to verify the result.

The Amprobe may be "infinite" resistance with some protection diodes,
so things may not be linear.

Connect a good 1 uF film capacitor across the Amprobe input, set to mV
range, apply 100 mV or whatever, disconnect source, and watch it
discharge. Then you can compute internal resistance.

John

L

#### LabMonkey

Jan 1, 1970
0
Ran into some problems:

the meter has an auto off feature which I can't turn off (at least I
couldn't find anything in the instruction manual). So after 10 minutes
it turns itself off. So with the capacitor experiment this only rules
out internal resistance with time constants less than 10 minutes. If
the internal resistance were 300M and the capacitor was 0.9mF, then the
time constant would be 4.5 minutes - which is not what I saw. After ten
minutes it had decayed to 0.200mV instead of 0.230mV.

As for the other experiment, the electrometer I use to measure the
voltage on the 100M resistor has some problems. I consistently got out
that the experimental equivalent parallel resistance was greater than
Rx, which naturally can't be.

I will try and find an electrometer which is more reliable, but as it
is the weekend I will not be able to get my hands on anything until
Monday.

Thanks for all of the help LabMonkey

L

#### LabMonkey

Jan 1, 1970
0
I found out how to trick the meter in staying on: quickly press the
"hold" button twice and it resets the clock for the 10 minute
inactivity.

Only problem is, when I did that I didn't get an internal resistance of
1.4G. The time constant would have been about 20 minutes and I had it
running for 4.5 hours with it only decreasing to 160mV from 257mV.

So either the capacitors were doing something funny, or the internal
resistance some how changes between even higher resistors when on the
"mV" range.

I am still trying to get a hold of an electrometer

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